Evaluate the limit
$$\displaystyle\lim_{n \to \infty} \int_{0}^{\pi} \frac{\sin x}{1+\cos ^2(nx)} dx$$
Using property of definite integral $\int_{0}^{2a} f(x).dx=2\int_{0}^{a} f(x)dx$,when $f(2a-x)=f(x)$ I got
$$\displaystyle\lim_{n \to \infty} \int_{o}^{\pi} \frac{\sin x}{1+\cos ^2(nx)} dx=2\displaystyle\lim_{n \to \infty} \int_{o}^{\pi/2} \frac{\sin x}{1+\cos ^2(nx)} dx$$ but I cannot proceed after that. Could someone provide me with some hint? Till now I have only done integration in terms of elementary functions. Any hint would be appreciated.
Best Answer
In this answer it is proved that if $f, g$ are Riemann integrable on $[0,T]$ and $g$ is periodic with period $T$ then $$\lim_{n\to\infty} \int_{0}^{T}f(x)g(nx)\,dx=\frac{1}{T}\left(\int_{0}^{T}f(x)\,dx\right)\left(\int_{0}^{T}g(x)\,dx\right)$$ For your problem we have $$f(x) =\sin x, g(x) =\frac{1}{1+\cos^2x}$$ and hence the desired limit is $$\frac{1}{\pi}\int_{0}^{\pi}\sin x\, dx\int_{0}^{\pi}\frac{dx}{1+\cos^2x}=\frac{4}{\pi}\int_{0}^{\pi}\frac{dx}{3+\cos x}=\frac{4}{\pi}\cdot\frac{\pi}{2\sqrt{2}}=\sqrt{2}$$